FP2 January 2006 Q3
3.
(a) Show that the substitution \(y = vx\) transforms the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x - 4y}{4x + 3y} \qquad \text{(I)}\] into the differential equation \[x\frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{3v^2 + 8v - 3}{3v + 4} \qquad \text{(II)}.\] (4)
(b) By solving differential equation (II), find a general solution of differential equation (I). (5)
(c) Given that \(y = 7\) at \(x = 1\), show that the particular solution of differential equation (I) can be written as \[(3y - x)(y + 3x) = 200.\] (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | B1 |
| \(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{3x - 4vx}{4x + 3vx}\) (all in terms of \(v\) and \(x\)) | M1 |
| \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{3 - 4v - v(4 + 3v)}{4 + 3v}\) (Requires \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{f}(v)\), 2 terms over common denom.) | M1 |
| \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\dfrac{3v^2 + 8v - 3}{3v + 4}\) | A1 cso |
| (4) |
Notes
(corrected from the printed mark scheme: the last line is printed without the minus sign, as \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{3v^2 + 8v - 3}{3v + 4}\))
| Scheme | Marks |
|---|---|
| \(\dfrac{3v + 4}{3v^2 + 8v - 3}\,\mathrm{d}v = -\dfrac{1}{x}\,\mathrm{d}x\) Separating variables | M1 |
| \(\pm\ln x\) | B1 |
| \(\dfrac{1}{2}\ln(3v^2 + 8v - 3)\) M: \(k\ln(3v^2 + 8v - 3)\) | M1 A1 |
| \(\dfrac{1}{2}\ln\left(\dfrac{3y^2}{x^2} + \dfrac{8y}{x} - 3\right) = -\ln x + C\) Or any equivalent form | A1 |
| (5) |
Notes
Parts (b) and (c) may well merge.
Partial fractions may be used \(\left(A = \dfrac{3}{2}, B = \dfrac{1}{2}\right)\), giving \(\dfrac{1}{2}\ln(3v - 1) + \dfrac{1}{2}\ln(v + 3)\).
| Scheme | Marks |
|---|---|
| \(\dfrac{3y^2}{x^2} + \dfrac{8y}{x} - 3 = \dfrac{A}{x^2}\) | |
| Removing ln’s correctly at any stage, dep. on having \(C\). | M1 |
| Using \((1, 7)\) to form an equation in \(A\) (need not be \(A = \ldots\)) | M1 |
| \((1, 7) \quad \Rightarrow \quad 3 \times 49 + 56 - 3 = A \quad \Rightarrow \quad A = 200\) (or equiv., can still be ln) | A1 |
| \(3y^2 + 8yx - 3x^2 = 200\) | |
| \((3y - x)(y + 3x) = 200\) (M dependent on the 2 previous M’s) | M1 A1 cso |
| (5) | |
| (14 marks) |
Notes
Final M requires formation and factorisation of the quadratic.